Solving Stoichiometry Problems with Gas Laws (PV=nRT) and Enthalpy (ΔH)

Stoichiometry is the accountant of chemistry. It tells us exactly how much reactant we need to produce a desired amount of product. However, reactions rarely happen in a vacuum—or, ironically, sometimes they do. When dealing with gases or energy changes, simple mass-to-mass conversions aren't enough.

You need to bridge the gap between the macroscopic world (pressure, volume, temperature, heat) and the microscopic world (moles). This guide covers advanced stoichiometry techniques, integrating the Ideal Gas Law ($PV=nRT$) and Enthalpy ($\Delta H$) to solve complex reaction problems.

The Ideal Gas Law (PV=nRT) in Stoichiometry

When substances react in the gas phase, we cannot weigh them on a balance scale. Instead, we measure their physical properties: Pressure ($P$), Volume ($V$), and Temperature ($T$). The bridge between these properties and the moles ($n$) required for stoichiometric ratios is the Ideal Gas Law:

Where $R$ is the ideal gas constant (typically $0.08206 \text{ L}\cdot\text{atm}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$).

Strategy for Gas Problems

1. Isolate Moles ($n$): Rearrange the equation to $n = \frac{PV}{RT}$.
2. Watch Your Units:
   • Pressure must match the $R$ value (usually atm).
   • Volume in Liters (L).
   • Temperature must be in Kelvin ($K = ^\circ C + 273.15$).
3. Apply Stoichiometry: Use the balanced chemical equation to convert moles of gas to moles of the target substance.

Illustrative Example: The Sodium Bicarbonate Balloon

Problem: You inflate a balloon using the reaction of solid sodium bicarbonate ($NaHCO_3$) with excess acid. The balloon is a sphere with a diameter of 12 inches. The room pressure is 0.998 atm and the temperature is 32°C. How much solid $NaHCO_3$ was used?

The Reaction: $NaHCO_3(s) + H^+(aq) \rightarrow Na^+(aq) + H_2O(l) + CO_2(g)$

Step 1: Calculate the Volume of Gas ($CO_2$)
First, convert the physical dimensions to metric volume.
Radius $r = 6 \text{ inches} = 15.24 \text{ cm}$.
Volume of a sphere $V = \frac{4}{3}\pi r^3 \approx 14,826 \text{ cm}^3 = 14.826 \text{ L}$.

Step 2: Solve for Moles of Gas ($n$)
Convert temperature to Kelvin: $32^\circ C + 273.15 = 305.15 \text{ K}$.
Using $n = \frac{PV}{RT}$: $$n_{CO_2} = \frac{(0.998 \text{ atm})(14.826 \text{ L})}{(0.08206 \text{ L}\cdot\text{atm}\cdot\text{mol}^{-1}\cdot\text{K}^{-1})(305.15 \text{ K})} \approx 0.591 \text{ mol}$$

Step 3: Stoichiometric Conversion
The reaction shows a 1:1 ratio between $NaHCO_3$ and $CO_2$. Therefore, we needed 0.591 moles of $NaHCO_3$.
Mass = $0.591 \text{ mol} \times 84.01 \text{ g/mol} \approx \mathbf{49.6 \text{ g}}$.

Gas Stoichiometry at Standard Temperature and Pressure (STP)

If a problem specifies STP (Standard Temperature and Pressure), calculations become significantly faster. Standard conditions are often defined as $T = 273.15 \text{ K}$ ($0^\circ C$) and $P = 1 \text{ atm}$.

Under these specific conditions, 1 mole of any ideal gas occupies 22.4 Liters.

Shortcut: Avogadro’s Law and Volume Ratios

If reactants and products are both gases at the same $T$ and $P$, their volume ratios are identical to their mole ratios. You do not need to calculate moles explicitly.

Practice Problem: Limiting Reagents with Gas Volumes

Problem: Consider the reaction: $3CO(g) + 7H_2(g) \rightarrow C_3H_8(g) + 3H_2O(g)$. If 27.0 L of $CO$ and 20.0 L of $H_2$ react at constant $T$ and $P$, which gas is in excess and by how much?

Solution: Since $P$ and $T$ are constant, we treat Liters just like Moles.

• Ratio Required: For every 3 L of $CO$, we need 7 L of $H_2$. ($Ratio = \frac{7}{3} \approx 2.33$)
• Ratio Available: $\frac{20.0 \text{ L } H_2}{27.0 \text{ L } CO} \approx 0.74$

Because the available ratio (0.74) is less than the required ratio (2.33), $H_2$ runs out first. It is the limiting reagent.

Calculate Excess $CO$:
Volume of $CO$ consumed = $20.0 \text{ L } H_2 \times \frac{3 \text{ L } CO}{7 \text{ L } H_2} = 8.57 \text{ L}$.
Remaining $CO$ = $27.0 \text{ L (initial)} - 8.57 \text{ L (consumed)} = \mathbf{18.4 \text{ L}}$.

Calculating Enthalpy Changes ($\Delta H$) for Reactions

Chemical reactions involve the transfer of heat. The Enthalpy of Reaction ($\Delta H_{rxn}$) is the heat absorbed or released during a reaction at constant pressure. It is an extensive property, meaning it depends on the amount of substance reacting.

Step-by-Step Case Study: Exothermic Reactions

Problem: The reaction of Carbon Disulfide with Chlorine is exothermic:
$$CS_2(g) + 3Cl_2(g) \rightarrow CCl_4(g) + S_2Cl_2(g) \quad \Delta H = -232.1 \text{ kJ}$$ How much energy is released if 11.9 g of $CS_2$ reacts?

1. Convert Mass to Moles:
   Molar Mass of $CS_2 = 12.01 + 2(32.07) = 76.15 \text{ g/mol}$.
   $$n = \frac{11.9 \text{ g}}{76.15 \text{ g/mol}} \approx 0.1563 \text{ mol } CS_2$$
2. Scale the Enthalpy:
   The equation says 1 mole of $CS_2$ releases 232.1 kJ. We have 0.1563 moles.
   $$\text{Energy} = 0.1563 \text{ mol} \times \frac{-232.1 \text{ kJ}}{1 \text{ mol } CS_2} = -36.28 \text{ kJ}$$
3. Interpretation:
   The negative sign indicates heat is released. The answer is 36.3 kJ released.

Quick Check: Sulfur Combustion

If $2S + 3O_2 \rightarrow 2SO_3$ has $\Delta H = -791.4 \text{ kJ}$, what is the enthalpy change for 6.44 g of Sulfur?

• Moles of S = $6.44 \text{ g} / 32.06 \text{ g/mol} = 0.201 \text{ mol}$.
• Crucial Step: The $\Delta H$ corresponds to 2 moles of S (see coefficient).
• $\Delta H_{actual} = 0.201 \text{ mol S} \times \frac{-791.4 \text{ kJ}}{2 \text{ mol S}} \approx \mathbf{-79.5 \text{ kJ}}$.

Hess's Law and Standard Enthalpies of Formation ($\Delta H_f^\circ$)

Sometimes you cannot measure the enthalpy of a reaction directly. Hess’s Law states that the total enthalpy change for a reaction is the same, regardless of the route taken. This allows us to calculate $\Delta H_{rxn}$ using known standard enthalpies of formation ($\Delta H_f^\circ$).

The standard formula is:

Example: Calculating Reaction Enthalpy from Formation Data

Goal: Calculate $\Delta H$ for: $CH_4(g) + 4Cl_2(g) \rightarrow CCl_4(g) + 4HCl(g)$

Given Data (Formation Reactions):

1. $C(s) + 2H_2(g) \rightarrow CH_4(g) \quad \Delta H_f^\circ = -74.6 \text{ kJ/mol}$
2. $C(s) + 2Cl_2(g) \rightarrow CCl_4(g) \quad \Delta H_f^\circ = -95.7 \text{ kJ/mol}$
3. $H_2(g) + Cl_2(g) \rightarrow 2HCl(g) \quad \Delta H = -184.6 \text{ kJ}$
   (Note: This produces 2 moles. $\Delta H_f^\circ$ for 1 mole HCl is $-184.6 / 2 = -92.3 \text{ kJ/mol}$)
4. $\Delta H_f^\circ$ for elements like $Cl_2(g)$ is always 0 kJ/mol.

Calculation:

Sum of Products:
$[1 \times \Delta H_f^\circ(CCl_4)] + [4 \times \Delta H_f^\circ(HCl)]$
$= [1(-95.7)] + [4(-92.3)] = -95.7 - 369.2 = -464.9 \text{ kJ}$

Sum of Reactants:
$[1 \times \Delta H_f^\circ(CH_4)] + [4 \times \Delta H_f^\circ(Cl_2)]$
$= [1(-74.6)] + [4(0)] = -74.6 \text{ kJ}$

Final $\Delta H_{rxn}$:
$\Delta H_{rxn} = (\text{Products}) - (\text{Reactants})$
$\Delta H_{rxn} = -464.9 - (-74.6) = -464.9 + 74.6 = \mathbf{-390.3 \text{ kJ}}$

Key Takeaways: Mastering Stoichiometry

• The Mole is the Center: Whether starting with grams, Liters (gas), or Energy (kJ), your first step is almost always to find moles.
• PV=nRT is the Key: Use the ideal gas law to convert between Gas Volume/Pressure and Moles. Remember: Temperature must be in Kelvin.
• Check Coefficients: When calculating Energy ($\Delta H$) or using volume ratios, always pay attention to the stoichiometric coefficients in the balanced equation. Energy is per "reaction event" as written.
• Hess's Law: Reaction energies are additive. If you reverse a reaction, flip the sign of $\Delta H$. If you multiply the reaction coefficients, multiply $\Delta H$ by the same factor.